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  • optointerrupter specs

    Ok, electronic question if this is an electronics forum. (not spam)

    I am trying to determine the minimum forward current (If) required in order to forward bias the phototransistor of an opto interrupter model H301-05.

    Data sheet:
    https://www.mouser.com/ds/2/239/H301-05-1141649.pdf

    Data sheet lists absolute max If of 60ma, and a If of 20ma at a Vf of 1.2v. But I see no mention of minimum current threshold required to switch on phototransistor.

    Thanks, John

  • #2
    As there is no minimum response time I would assume that these specs are under continuous operation. Only the OEM can tell you the info not listed.
    ~~

    ... it was plumbed by Ray Charles and his helper Stevie Wonder

    Comment


    • johncameron
      johncameron commented
      Editing a comment
      There is a response time of about 30microseconds with 2k load but regardless the minimum If (tr) of ir diode to bias transistor is not specifically mentioned.

  • #3
    What would this be typically be used for?

    Comment


    • Bob D.
      Bob D. commented
      Editing a comment
      An optical interrupter circuit can be used in an alarm system or in manufacturing to count items traveling on an assembly line to give a couple examples. A light beam is transmitted between the emitter and receiver., when the beam is broken (interrupted) by an object then an alarm can be triggered just as if a switch had been open or closed.

      A common application of this would be the safety circuit on a garage door opener. You have the sensors (one transmitter and one receiver) installed down near the ground so that if for example a child or small animal or object is in the path of the door when it is closing the doors movement will be stopped and reversed.

      Or say you have bottles of beer racing down an assembly line and you need to count how many are being filled per hour. As each bottle passes between the sensors and interrupts the beam the counter is increased by one. No physical contact between the bottle and the sensor required. Some security alarms systems use a circuit of this type too. Then can be set to alarm on a make (beam established) or break (beam interrupted) condition, just like a NO or NC switch.

      But I think the item John is talking about is a self-contained device, meaning both the transmitter and receiver are contained in the same IC package with a gap in between. Like an opt-isolator but with a physical gap between that would allow some external physical item to break the beam and trigger the device.

      An opto-isolator IIRC is used when you need electrical separation between two circuits, no current flow between them, but need to transmit data or a signal of some sort. Used in data communications and many other commercial and industrial applications.

  • #4
    Excellent explanation BobD, the term sort stuck in my head and I was curious about the story behind it.

    Comment


    • #5
      My brother in law's first job after graduating with an EE was working on a beam to identify whether or not the prize fell into the box of Cracker Jacks.

      Comment


      • #6
        I replaced an opener motor once and couldn't understand why it wouldn't work properly. My EE father in law, who was a master trouble shooter, by inclination as well as training, spent an hour running the door through it's hoops, never using intuition, just going a to z, as he always did when trying to solve a problem. He finally came up with the solution, at which point, I about fell over with embarrassment. When I attached the chain to the motor, I didn't run the gear all the way to the end and when engaged the door would only go part way down. A simple fix, perhaps, as are most after the solution.
        Last edited by Plumbus; 11-16-2018, 03:27 PM.

        Comment


        • Mightyservant
          Mightyservant commented
          Editing a comment
          That never happens to me,....no never

      • #7
        According to Fig 2 in the datasheet the diode turns on when the voltage across it is about 1V. From 0V - 1V Rd will not conduct and so there will be 0V PD across it. As the voltage across the diode reaches 1V current will start flowing and from there on raising the input voltage will increase current through the diode. So you can think of the turn on of the diode to be controlled by the input voltage rather than current. The input voltage needs to be > 1V to start the diode conducting. When turned on you need to make sure that the source can supply the worst case current to keep it turned on. It's not clear from the datasheet what the nominal voltage across the diode will be when the diode is biased through Rd. Most general purpose diodes will be around 0.7V. It looks like for the purposes of design here you can use the typical value of 1.2V mentioned in the datasheet.

        Comment


        • #8
          Thanks but it doesn't answer my question, although I've pretty much figured out what I need to know.

          Photodiodes act similarly to other leds expect the forward voltage is much lower.

          Just saying "turn on the diode" means nothing. It will emit light even in the microamp range. What I was asking was what is necessary to switch on the photo transistor.
          That's more complicated and has to do with how the phototransistor is connected.

          It has to do with the collector emitter voltage vs current and how its loaded. This is similar to the current transfer ratio used in similar optocouplers. (Ic=CTRxIf LED)

          Page 4 fig three of datasheet shows a comparison graph showing this relationship of switching characteristics of the transistor and how it relates to the forward current of the diode.

          So with only @4ma forward current through the photo diode I should have about 0.4ma current through the collector of the transistor given a 1v collector voltage.

          That's about as close an answer as I need.

          Sizing the current limiting resistor is obviously very important. This photodiode can not see anything over 60ma or it would burn out instantly, but running it at its lowest current in order to do its job to bias the transistor is desirable because it will dramatically increase its lifespan.

          So long story short, 3 or 4 ma is about its minimum diode forward current in most applications.

          Comment


          • #9
            Ah ok - I guess I was not clear on your original question.

            From what I can tell based on Fig 2 and Fig 3 you can pick your operating point as you see fit as described by the curves. It's not clear exactly what current in the diode circuit will cause the transistor to become active. The transistor could well turn on below an If of 4mA - the info is just not available from Fig 3. But picking an operating point of 5mA would be safe. I do not know the rest of the circuit where you are planning to use this in order to understand the benefit of going with a higher Ic.

            Comment


            • #10
              Thanks Blue_can;

              Exactly, no. Fig 3 graph is just too small and crude to read accurately.But the info is there.

              The amount the transistor is turned on is read vertically in collector current, the Vce across the transistor is read horizontally, and the If of the led is written in the curves of the graph.
              If you know all three you have a good idea of how to set up the circuit.

              Given that a typical Vf of this led is listed at 1.2v the current should be around 20ma. But it's operating range is about 3 to 60ma.

              In my repair situation, I was trying to figure out an unknown zener value in the circuit based on the current through this PI. I extrapolated its value on the low end (4ma) of that range and all is working fine now.

              Thanks for your help.

              Comment


              • #11
                Glad everything worked out for you. I did not realize you were repairing something - I thought you were designing something that this was a part of so my answers were geared towards that. I believe the reason this info is not available in the datasheet is that it would be of little value to anyone trying to design with it since you would want to stay away from the "just turned on" point.

                The only way to get this answer is if fig 3 were to continue plotting reducing values of If until Ic flatlines at 0. I currently don't see that in the plot. The spread you see in the plot when you zoom into the datasheet for a given If is likely to represent the variance between parts due to manufacturing tolerances.

                The other way would be to have a plot of If vs Vbe which does not exist. For most typical BJTs Vbe needs to be around 0.7V in order to start Ic conducting.

                Another way to figure this out would be to build up a test harness with this part and then increase If while monitoring Ic. Again this would not be super accurate since there will be a spread of values based on the manufacturing tolerances but would have been good enough for what you wanted to know for your purpose.

                Comment


                • #12
                  Originally posted by blue_can View Post
                  The only way to get this answer is if fig 3 were to continue plotting reducing values....I currently don't see that in the plot. The spread you see in the plot when you zoom into the datasheet for a given If ...
                  Thanks. Yes, its not entirely plotted but you can definitely see the range, at least crudely.

                  I believe tolerance in manufacturing is shown in groupings of three for each forward current category of led current, but for each current category 5,10,15,20ma etc. you can see how it changes the collector current.

                  In hindsight the question was unanswerable because the point the transistor turns on can't be discerned without knowing the entire circuit. The way its supplied and loaded makes a big difference.

                  One example would be a 5v supply and a load of 10k. About 4v would be dropped across the resistor leaving a Vce of 1v. Now if the led had about 4 to 5ma through it, it would correlate with the graph I(SUB)C of 0.5ma. Which incidentally is the minimum on state collector current listed on page 3.

                  I agree that setting up a bench supply to test these points is often the best approach but I didn't understand the datasheet at first.
                  I was looking for a CTR when there isn't one.

                  Once I got a better understanding of the datasheet, the math worked out and I was able to discern the parts needed.

                  Comment


                  • #13
                    Yes you're correct that the question is not easy to answer with the data provided. As you say there is also the big picture - both in terms of analyzing the circuit for an existing system or when designing this into a new design. With complex circuits analysis could become difficult and sometimes the intent of the original designer may not be obvious.

                    Another approach to a bench setup would be to use SPICE to model the circuit - but you would need an accurate model of the parts.

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